Question

Need help .

A,B,C are the angles of a triangle, then
SinA + SinB - SinC = 4Sin(A/2) Sin(B/2) Cos(C/2)

Answer 1

what is the question...

Answer 2

prove it . thats all

Answer 3

C=180-A-B
sinA+sinB-sinC = sinA+sinB-sin(180-(A+B))
=sinA+sinB-sin(A+B)
=sinA+sinB-sinAcosB-sinBcosA
=sinA(1-cosB)+sinB(1-cosA)
=2sinAsin\(^{2}\)(B/2)+2sinBsin\(^{2}\)(A/2)
=4sin(A/2)sin\(^{2}\)(B/2)cos(A/2)+4sin(B/2)sin\(^{2}\)(A/2)cos(B/2)
=4sin(A/2)sin(B/2) [sin(B/2)cos(A/2)+sin(A/2)cos(B/2)]
=4sin(A/2)sin(B/2)sin(A/2+B/2)
=4sin(A/2)sin(B/2)sin(90-C/2)
=4sin(A/2)sin(B/2)cos(C/2)

Answer 4

hope that is readable